IMA Preprint Series # 2385

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1 LIST COLORINGS WITH DISTINCT LIST SIZES, THE CASE OF COMPLETE BIPARTITE GRAPHS By Zoltán Füredi and Ida Kantor IMA Preprint Series # 2385 ( October 2011 ) INSTITUTE FOR MATHEMATICS AND ITS APPLICATIONS UNIVERSITY OF MINNESOTA 400 Lind Hall 207 Church Street S.E. Minneapolis, Minnesota Phone: Fax: URL:

2 List colorings with distinct list sizes, the case of complete bipartite graphs Zoltán Füredi 1 Rényi Institute of the Hungarian Academy Budapest, P.O.Box 127, Hungary, H-1364, furedi@renyi.hu and Department of Mathematics University of Illinois at Urbana-Champaign Urbana, IL 61801, USA z-furedi@illinois.edu Ida Kantor Institute for Theoretical Computer Science Charles University Malostranské nám. 25, Praha 1, Czech Republic ida@kam.mff.cuni.cz Abstract Let f : V N be a function on the vertex set of the graph G = (V, E). The graph G is f-choosable if for every collection of lists with list sizes specified by f there is a proper coloring using colors from the lists. The sum choice number, χ sc (G), is the minimum of f(v), over all functions f such that G is f-choosable. It is known (Alon 1993, 2000) that if G has average degree d, then the usual choice number χ l (G) is at least Ω(log d), so they grow simultaneously. In this paper we show that χ sc (G)/ V (G) can be bounded while the minimum degree δ min (G). Our main tool is to give tight estimates for the sum choice number of the unbalanced complete bipartite graph K a,q. Key words and Phrases: graphs, list chromatic number Mathematics Subject Classification: 05C15, 05C35, 05C65. [furedi_kantor_sum_choice_2011_oct_31.tex] Printed on October 31, Research supported in part by the Hungarian National Science Foundation OTKA, by the National Science Foundation grant DMS , and by a European Research Council Advanced Investigators Grant Acknowledgement: This research was supported in part by the Institute for Mathematics and its Applications with funds provided by the National Science Foundation. A preliminary version was presented in: European Conference on Combinatorics, Graph Theory and Applications (EuroComb 2009), , Electron. Notes Discrete Math. 34, Elsevier Sci. B. V., Amsterdam,

3 1 Average list sizes and planar graphs Given a graph G and a list of colors L(v) for each vertex v V (G), we say that G is L-choosable (or that L is sufficient) if it is possible to choose c(v) L(v) for all v so that c : V (G) L(v) is a proper coloring of G. The choice number (or list chromatic number) χ l is the minimum t such that for every assignment L with L(v) t for all v V, the graph is L-choosable. It is well-known (Thomassen [8]) that χ l (P ) 5 (1) for every planar graph P, and this is the best possible ([9]). However, if we allow distinct list sizes, then the average size can be smaller. For example, Thomassen s beautiful proof for (1) gives that if P is an n-vertex planar graph, v 1,..., v t are its external vertices (in this order) and the list sizes are 1 for v = v 1, 2 for v = v L(v) = 2, (2) 3 for v = v 3,..., v t, 5 for the inner vertices, then P is L-choosable. Consider a function f : V (G) N. An f-assignment is an assignment of lists L(v) to the vertices v V (G) such that L(v) = f(v) for all v. The function f is sufficient if G is L-choosable for all f-assignments L. We define the sum choice number of G, denoted by χ sc (G), as the minimum of v V (G) f(v) over all sufficient f. Sum choice numbers were introduced by Isaak in [6] who proved that if G is the line-graph of K 2,q then χ sc (G) = q 2 + 5q/3. Various classes of graphs were investigated by Isaak in [7], by Berliner, Bostelmann, Brualdi, and Deaett [3] and by Heinold in [4] and [5]. Thomassen s theorem (2) implies that χ sc (P ) 5n 9 for planer P (n 2). In fact, more is true. It is easy to show (see, e.g., [7]) that for every graph χ sc (G) V (G) + E(G) (3) holds. Hence χ sc (P ) 4n 6. Our first result is a slight improvement. Theorem 1. Let P be an n-vertex planar graph. There exists an f : V (P ) N such that f(v) 4n 6, max f(v) 6, and P is f-choosable. Proof. Consider a linear order of the vertices of P, and let ˆd(v) be the number of neighbors of v that precede it. The function f(v) = ˆd(v) + 1 is a sufficient function, so ( ˆd(v) + 1) yields an upper bound on χ sc (P ). Since every planar graph has a vertex of degree at most 5, it is possible to order the vertices so that ˆd(v) 5 for all v.

4 Z. Füredi and I. Kantor: Sum-choice 2 2 Unbalanced complete bipartite graphs Erdős, Rubin and Taylor (see, e.g., [1]) showed for the complete bipartite graph that χ l (K q,q ) = Θ(log q). (4) If one of the parts is substantially smaller than the other one, then allowing different list sizes results in smaller average lists. It is easy to show χ sc (K 1,q ) = 2q + 1 (as for every tree on q + 1 vertices). Berliner, Bostelmann, Brualdi, and Deaettet [3] showed that for all q 1 we have χ sc (K 2,q ) = 2q q + 1, (5) and Heinold [5] proved χ sc (K 3,q ) = 2q q + 4. (6) Our main result deals with the sum choice number of K a,q with arbitrary a. Theorem 2. There exist positive constants c 1 and c 2 such that for all a 2 and q 4a 2 log a 2q + c 1 a q log a χ sc (K a,q ) 2q + c 2 a q log a. It is known that χ l is not independent of the average degree. Alon [1, 2] proved that for some constant c > 0, every graph G with average degree d has χ l (G) c log d. (7) An easy corollary of our Theorem 2 is that if different list sizes are allowed, then such dependence does not exist. Indeed, we have 2 E(K a,q ) lim a, a + q q>>a 2 log a χ sc (K a,q ) =, a, lim = 2. a + q q>>a 2 log a So the structure of the graph plays a more important role in determining the sum choice number than in the case of the list chromatic number. 3 Upper bound, there are sufficient short lists Throughout this paper, the two parts of the complete bipartite graph K a,q will be denoted by A and Q, with A = a and Q = q. Theorem 3. Suppose that a, q N with q a 2. Then χ sc (K a,q ) 2q + a 32q(1 + log a). Proof. To prove the upper bound, we present a function f with v A Q f(v) = 2q + a 32q(1 + log a) such that every f-assignment is sufficient.

5 Z. Füredi and I. Kantor: Sum-choice 3 Define f as f(v) = { r for v A; 2 for v Q where r will be defined later in (10) as any integer r 32q(1 + log a). Let L be an arbitrary f-assignment, i.e., L(v) = f(v) for all v. Consider S := v A Q L(v). The assignment L yields a (multi)hypergraph and a multigraph on the same vertex set S and with edge sets L A := {L(u) : u A} and L Q := {L(v) : v Q}, respectively. Sufficiency of L means that one can find a set T S meeting all hyperedges of L A such that S \ T meets all edges of L Q, so T is an independent set in the graph L Q. Given T the choice function c can be defined as and c(u) L(u) T, for u A c(v) L(v) (S \ T ), for v Q. We are going to construct such T by a 2-step random process. Let us pick, randomly and independently, each element of S with probability p. Let B be the random set of all elements picked. Define a random variable X u for each u A as X u = L(u) B, and the random variable Y by Y := {v Q : L(v) B}, so Y is the number of edges of L Q spanned by B. Remove an element c(v) L(v) for each edge of L Q spanned by B, the remaining set T B is certainly independent in L Q, and if Y < X u for each u A, then T meets all L(u) L A and we are done. The expected value of Y is p 2 q, so Markov inequality gives The expected value of X u is pr, so Chernoff inequality gives for any t > 0. Hence Prob (Y < 2p 2 q) 1 2. (8) Prob (X u < EX u t) < e t2 /2rp, Prob (X u pr t for all u A) > 1 ae t2 /2rp, (9) and this is at least 1/2 for t 2 2rp ln(2a). The sum of probabilities in (8) and (9) is larger than 1, so there is an appropriate choice of B (and then T ) if t 2 = 2rp(1 + log a) and pr t 2p 2 q. We can define, e.g., 2(1 + log a) p := and r 4pq = 32(1 + log a)q. (10) q

6 Z. Füredi and I. Kantor: Sum-choice 4 4 Lower bound, much shorter lists are not sufficient To prove that χ sc (G) k for a particular k, we need to show that for every f with v G f(v) = k, there exists an insufficient f-assignment. First, we show how to construct an insufficient assignment for some special f. Lemma 1. Let t 2 and l 1. assignment L with L(v) = For a = 2 t and q = tl 2, there exists an insufficient { tl for v A, 2 for v Q. Proof. Take 2t pairwise disjoint sets X i, Y i of size l, Z = (X i Y i ). Identify the elements of A by the set of 0-1 vectors of length t, A = {0, 1} t. For a vector (ε 1,..., ε t ) A define L(v) as ( ε i =1 X i) ( ε j =0 Y j). So L(v) contains either X i or Y i for all i. Let the graph G be the union of t complete bipartite graphs on the vertex set Z by setting E(G) = t i=1 (X i Y i ) and define the lists L(v) for v Q as the edges of G. The number of edges of G is tl 2 = q, so a one-to-one mapping can be done. Every independent set T of G contains at most one vertex from each X i Y i so it cannot meet all hyperedges of L A, where L A := {L(v) : v A}. This means that this assignment L is not sufficient. Note that with this choice of a and q, we have L(v) = q log 2 a for v A. Also notice that if we remove some elements from the lists in the above construction, the resulting list assignment is still insufficient. Theorem 4. If a 2 and q > 4a 2 log a, then χ sc (K a,q ) 2q a q log a Proof. Suppose that f : V (K a,q ) N with v A Q f(v) = 2q + as where s 0.06 q log a. We will find an insufficient f-assignment. Let q 1, q 2 and q 3 be the numbers of vertices v Q with f(v) = 1, f(v) = 2 and f(v) 3, respectively. If f(u) q 1 for some u A, then f is obviously insufficient. From now on, we suppose that f(u) > q 1 for each u A. We obtain 2q + as = f(v) aq 1 + (q 1 + 2q 2 + 3q 3 ) 2q + q 1 + q 3. (11) v A Q It follows that q 1 + q 3 as and Q has at least q as vertices with lists of size 2. Let q = q 2 and let a be the largest power of 2 not exceeding 1 2 a. If there are at least a vertices u A with f(u) q log 2 a, then we can use Lemma 1 to construct an insufficient assignment. If this does not hold, then A has more than a 2 vertices with lists of size greater than (q as) log2 a. Using a > a/4 we obtain v A Q f(v) 1 2 a (q as) log 2 (a/4) + 2q as. (12)

7 Z. Füredi and I. Kantor: Sum-choice 5 The rest is a little calculation to show that here the right hand side exceeds 2q +0.06a q log a for a 5, q > 4a 2 log a and s < 0.06 q log a. Finally, the case a 4 (in fact a 30) follow from (5) and (6), completing the proof. Let us remark that if we choose the constants in the proof of Theorem 3 more carefully, we can improve the constant 32 to Using a randomized construction, it is possible to improve the constant 0.06 in Theorem 4 to For fixed a, a limit exists as q In this section we suppose that a 2 is a fixed integer. We have proved bounds for α q := (χ sc (K a,q ) 2q)/ q. Now we show that in fact the limit exists when q tends to. Theorem 5. For fixed a, the limit lim q χ sc (K a,q ) 2q q exists. First, we consider a simpler problem and consider only type II assignments of K a,q which means f(v) = 2 for all v Q. Define χ sc2 (K a,q ) to be the minimum of A Q f(v) where f runs over all sufficient type II functions. Obviously χ sc2 (K a,q ) χ sc (K a,q ). Theorem 6. For fixed a, the limit lim q χ sc2 (K a,q ) 2q q exists. A type II f is not sufficient if and only if there exists a a hypergraph L with edges L i satisfying L i = f i for i = 1,..., a and a graph G on V (L) with at most q edges, such that no transversal of L is an independent set in G. For I [a], define X I = i I L i. An insufficient II assignment is symmetric if for all pairs I J, the bipartite subgraph of G induced by X I and X J is either empty of complete, and for each I, X I induces the empty graph. Without loss of generality we may assume that an insufficient type II assignment is symmetric, as the following lemma demonstrates. From now on, in this section, all assignments are of type II, except when stated otherwise. Lemma 2. Given a, q and f an insufficient type II assignment exists if and only if a symmetric insufficient assignment exists. Proof. Suppose that (L, G) is an insufficient assignment. If u and v belong to the same X I, then no minimal transversal of L contains both of them. We can therefore delete all edges induced by X I. Now suppose that u, v X I and N(u) N(v). Replace the neighborhood of v by the neighborhood of u. It is still true that every transversal of L induces an edge of G. Repeated application of this procedure eventually produces a symmetric insufficient assignment. Proof of Theorem 6. Consider a symmetric insufficient assignment f for K a,q. Let L, G and X I (for I [a]) be as before, x I = X I. Let V := {v I : I [a]} be a 2 a -element set. Let R be the reduced graph of the symmetric insufficient assignment, i.e., the graph with V (R) = {v I ; x I 0} and whose edges correspond to the complete bipartite subgraphs of G. Similarly, the hypergraph L turns into the reduced hypergraph on the same vertex set, V (R).

8 Z. Füredi and I. Kantor: Sum-choice 6 The graph R is blocking i.e., every vertex cover of the reduced hypergraph contains an edge of R. The vector x = (x I ) satisfies IJ E(R) x Ix J q, and x I = 0 whenever v I V (R). The set A q R of all such x lies in the non-negative orthant of R2[a] and is bounded by a quadric surface which depends on R and q. Define the linear map ϕ : R 2[a] R a by ϕ(x) = (f 1,..., f a ) where f i = i I x I. The function f is insufficient for this q if and only if f is the image of some integer point x that is in A q R for some blocking R. If there exists an insufficient f-assignment for every integer vector f such that f i = k, then we have χ sc2 (K a,q ) 2q > k. We are therefore looking for the maximum k such that every integer point on the hyperplane f i = k is the image (under ϕ) of some integer point in A q R, where the union is taken over all (but finitely many) blocking R s. Let us normalize everything by q. For every blocking R, define A R := {x : IJ E(R) x I x J 1, and x I = 0 for v I V (R)} and B R := ϕ(a R ). For every R we now have only one quadric surface, independent of q. We say that a vector v is a q-grid point if q v is an integer point. For every q, define k q to be the maximum k such that every q-grid point on the hyperplane fi = k is the image of some q-grid point in A R. Also, define β to be the maximum k such that the simplex C k := {f : f i k} is a subset of B R. We want to prove that the limit lim k q exists and equals β. That is, we want to prove that for every ε, if q is large enough, every q-grid point in C β ε is the image under ϕ of some q-grid point in A R, and there is a q-grid point in C β+ε which is not the image of any q-grid point in A R. To prove the first claim, fix q and let f be a point on the hyperplane f i = β. The point f is in B R, so it is the image of some x A R. Each set A R is a downset in the sense that with every x it also contains all points z such that z i x i for all i. It follows that y := q x q is a q-grid point in A R. Each entry of y differs by at most 1 q from the corresponding entry of x, and a simple computation suffices to show that the distance of ϕ(y) and f is at most c q, where c is a constant dependent only on a. That is, for each point f on the hyperplane fi = β we have found, in distance at most ε q := c q, an image of a q-grid point from A R. Call this point f. Note that, by definition of χ sc, whenever a q-grid point f is the image of a q-grid point in A R, the same is true for all q-grid points in the box D f := {g : g i f i for all i}. Let h be a q-grid point such that h i β ε q a. Let f be its perpendicular projection on the hyperplane f i = β and find the corresponding f. Since the distance of f and f is at most ε q, the point h belongs to D f, and hence it is the image of a q-grid point in A R. Choosing q large enough so that ε q a ε for our given ε concludes the proof.

9 Z. Füredi and I. Kantor: Sum-choice 7 Now we prove the second claim. Let f be a point outside B R, but within the distance ε from C β. Take a bounded cube Q R a that contains f. Now take a bounded cube S in R 2a which contains all points x such that ϕ(x) Q. Then T := S ( A R ) is a compact set, so ϕ maps it to a compact set. The complement of ϕ(t ) in ϕ(s) is open, and contains f. Note that ( B R ) Q ϕ(t ). Therefore, for some small δ, the δ-ball around f is outside B R. If q is large enough, the ball contains some q-grid point. This point not only has no q-grid preimages in A R, it has no preimages in A R whatsoever, and the claim is proven. Proof of Theorem 5. Define sequences {α q } q=1 and {β q} q=1 as follows α q = χ sc(a, q) 2q q and β q = χ sc2(a, q) 2q q. It was already mentioned that α q β q for all q. The argument in the proof of Theorem 4 shows that whenever we have an insufficient function f for K a,q, we can delete at most d(q) = O( q) vertices of Q and get an insufficient function for K a,q d(q) where f(v) = 2 for v Q. We therefore have α q q = χsc (a, q) 2q χ sc2 (q, q d(q)) 2(q d(q)) = β q d(q) q d(q). We get the following relationship between α q and β q : q d(q) β q α q β q q d(q). The limit lim q β q exits by Theorem 6. Since d(q) = O( q), we have lim q q d(q) β q q d(q) = lim β q, q which proves the claim. 6 Graphs with large independent sets and a generalization of Turán s theorem Let G a,q be the graph that we get from K a,q by inserting an edge {u, v} for every pair of distinct u, v A. Theorem 7. There exist positive constants c 1 and c 2, independent of q and a, such that for q > a 2 we have 2q + c 1 a q(a 1) χ sc (G a,q ) 2q + c 2 a q(a 1). We will use the following generalization of Turán theorem. For given positive integers s and k, let t(s, k) = min ( di ) 1 i k 2, where the minimum is taken over all non-negative integer sequences (x 1,..., x k ) such that d i = s.

10 Z. Füredi and I. Kantor: Sum-choice 8 Theorem 8. Let s, a 2 be integers and let G be a graph with less than t(s, a 1) edges. Let L 1,..., L a V (G) be sets of size s. Then there exists a system of distinct representatives {u 1,..., u a } of L 1,..., L a, (i.e., u i L i ) which is an independent a-element set in G. The case V (G) = L 1 = = L a gives (the dual form of) Turán s theorem. Let us also remark that the result concerning E(G) is sharp: taking L 1 = = L a = V (G), V (G) = s with G being the disjoint union of a 1 cliques of almost equal sizes provide a graph of t(s, a 1) edges and a family without any system of distinct representatives which is independent in G (because G has no any independent set of size a). Proof. We define the sequence of distinct vertices u 1,..., u a one by one by an algorithm, such that u k L ik, where {i 1, i 2,..., i a } is a permutation of {1, 2,..., a} and also the set {u 1,..., u a } is independent in G. Let V 1 = L 1 L a and G 1 = G[V 1 ] (the restriction of G to V 1 ). Let u 1 be the vertex of minimum degree in G 1, D 1 the closed neighborhood of u 1 in G 1, and d 1 = D 1. Let L i1 be one of the hyperedges containing u 1. If u j, D j and L ij are already defined for j = 1,..., k, consider V k+1 = ( i {i 1,...,i k } L i) \ (D 1 D k ), G k+1 = G[V k+1 ], and let u k+1 be a vertex of minimum degree in G k+1, D k+1 its closed neighborhood in G k+1, d k+1 = D k+1, and L ik+1 one of the hyperedges containing u k+1, different from L i1,..., L ik. We claim that this algorithm only stops after a steps thus supplying the desired independent set {u 1,..., u a }. If we cannot define u k+1 for a k < a, then V k+1 is empty, and D D k s. As D 1,..., D k are non-empty, disjoint sets we get the contradiction 2 E(G) deg(u) d 1 (d 1 1) + + d k (d k 1) 2t(s, k) 2t(s, a 1). u D 1 D k Proof of Theorem 7. Again, we will prove the upper bound (with c 2 = 3) by presenting a sufficient function f. Let { s for v A, f(v) = 2 for v Q, where s := 3 (a 1)q. Note that s a and t(s, a 1) > q. Consider any f-assignment, it provides sets L 1,..., L a and a graph G with E(G) = q. This assignment is sufficient by Theorem 8. For the lower bound, we proceed as in Section 4 and first consider f s with f(v) = 2 for all v Q. Let {v 1,..., v a } be the vertices of A and let { si for v = v f(v) = i A and 2 for v Q such that s 1 s a. We claim that if f is sufficient, then q < t(s i, i 1) for all i 2. These inequalities imply that s i 2(i 1)q so f(v) 2q + i 1 2q 2q a (a 1)q. 1 i a

11 Z. Füredi and I. Kantor: Sum-choice 9 Suppose, on the contrary, that q t(s i, i 1) for some i 2. Let G be the graph consisting of i 1 disjoint cliques with sizes as equal as possible. Assign the pairs corresponding to the edges of G as the lists for the vertices of Q and let L(v 1 ) L(v i ) = V (G). This assignment is not sufficient. To finish the proof of the lower bound for an arbitrary sufficient f with f(v) = 2q + as we use the inequality (11) to obtain that f(v) = 2 for all but at most q as vertices v Q. Then we conclude the proof with an argument analogous to (12). The details are omitted. Problem 9. Suppose that { si for v = v f(v) = i A and 2 for v Q such that s 1 s a. What conditions are sufficient and necessary for f being G a,q -sufficient? We already have seen that q < t(s i, i 1) for all i 2 are necessary. It is easy to see that q ( s 12 ) + s1 (s 2 s 1 ) is also necessary, let G be the graph that we get by taking a clique of order s 2 and deleting edges of a clique of order s 2 s 1. Assign its edges as the lists of Q, and let L(v 1 ) L(v 2 ) = V (G). One is tempted to conjecture that these conditions altogether are already sufficient. This is the same problem as to ask that how much the sizes of L i s can be decreased in Theorem 8. References [1] N. Alon, Restricted colorings of graphs, Surveys in Combinatorics, 1993, pp [2] N. Alon, Degrees and choice numbers, Random Structures and Algorithms, 16 (2000), pp [3] A. Berliner, U. Bostelmann, R. A. Brualdi, and L. Deaett, Sum list coloring graphs, Graphs and Combin., 22 (2006), pp [4] B. Heinold, Sum list coloring and choosability, PhD thesis, Lehigh University, Bethlehem, [5] B. Heinold, Sum choice numbers of some graphs, Discrete Math., 309 (2009), pp [6] G. Isaak, Sum list coloring 2*n arrays, Electr. J. Combin., 9 (2002). [7] G. Isaak, Sum list coloring block graphs, Graphs and Combin., 20 (2004), pp [8] C. Thomassen, Every planar graph is 5-choosable, J. Combin. Theory Ser. B, 62 (1994), pp [9] M. Voigt, List colourings of planar graphs, Discrete Math., 120 (1993), pp